Question

A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long.

Answer 1

Answer:

74.86% probability that a component is at least 12 centimeters long.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 14$

Variance is 9.

The standard deviation is the square root of the variance.

So

$\sigma = \sqrt{9} = 3$

Calculate the probability that a component is at least 12 centimeters long.

This is 1 subtracted by the pvalue of Z when X = 12. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12 - 14}{3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

1-0.2514 = 0.7486

74.86% probability that a component is at least 12 centimeters long.